MCQMediumJEE 2023Conic Sections (Parabola, Ellipse, Hyperbola)

JEE Mathematics 2023 Question with Solution

Let PQPQ be a focal chord of the parabola y2=36xy^2 = 36x of length 100100, making an acute angle with the positive xx-axis. Let the ordinate of PP be positive and MM be the point on the line segment PQPQ such that PM:MQ=3:1PM : MQ = 3 : 1. Then which of the following points does NOT lie on the line passing through MM and perpendicular to the line PQPQ?

  • A

    (3,33)(3, 33)

  • B

    (6,29)(6, 29)

  • C

    (6,45)(-6, 45)

  • D

    (3,43)(-3, 43)

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The parabola is y2=36xy^2 = 36x, so in standard form y2=4axy^2 = 4ax we have a=9a = 9. Also, PQPQ is a focal chord of length 100100, the ordinate of PP is positive, and MM divides PQPQ in the ratio 3:13:1.

Find: Which given point does not lie on the line through MM perpendicular to PQPQ.

For the parabola y2=4axy^2 = 4ax, a parametric point is (at2,2at)\left(at^2, 2at\right). Hence here

P=(9t2,18t)P = \left(9t^2, 18t\right)

and since PQPQ is a focal chord, the other point is obtained using t1t2=1t_1 t_2 = -1, so

Q=(9(1t)2,18(1t))Q = \left(9\left(-\frac{1}{t}\right)^2, 18\left(-\frac{1}{t}\right)\right)

Using the given length condition PQ=100PQ = 100 and the fact that the ordinate of PP is positive, we get

P=(81,54),Q=(1,6)P = (81, 54), \quad Q = (1, -6)

Using section formula and line equation

Now MM divides PQPQ internally in the ratio PM:MQ=3:1PM:MQ = 3:1. Therefore,

M=P+34(QP)M = P + \frac{3}{4}(Q - P)

Substituting P=(81,54)P = (81, 54) and Q=(1,6)Q = (1, -6),

M=(81,54)+34((1,6)(81,54))M = (81, 54) + \frac{3}{4}\left((1, -6) - (81, 54)\right) M=(81,54)+34(80,60)M = (81, 54) + \frac{3}{4}(-80, -60) M=(81,54)+(60,45)=(21,9)M = (81, 54) + (-60, -45) = (21, 9)

Perpendicular line check

The slope of PQPQ is

54(6)811=6080=34\frac{54 - (-6)}{81 - 1} = \frac{60}{80} = \frac{3}{4}

So the slope of the perpendicular line is

43-\frac{4}{3}

Through M=(21,9)M = (21, 9), its equation is

y9=43(x21)y - 9 = -\frac{4}{3}(x - 21)

which simplifies to

y=43x+37y = -\frac{4}{3}x + 37

Now check the options:

  • For (3,33)(3, 33): y=43(3)+37=33y = -\frac{4}{3}(3) + 37 = 33, so it lies on the line.
  • For (6,29)(6, 29): y=43(6)+37=29y = -\frac{4}{3}(6) + 37 = 29, so it lies on the line.
  • For (6,45)(-6, 45): y=43(6)+37=45y = -\frac{4}{3}(-6) + 37 = 45, so it lies on the line.
  • For (3,43)(-3, 43): y=43(3)+37=4143y = -\frac{4}{3}(-3) + 37 = 41 \ne 43, so it does not lie on the line.

Therefore, the point that does not lie on the required line is (3,43)(-3, 43).

The solution states The Correct Option is C, but the actual point found from the working is (3,43)(-3, 43), which corresponds to option D in the listed options. Since the options do not match the solution label, the defensible answer from the working is D.

Common mistakes

  • Using the wrong focal chord condition. For a focal chord of y2=4axy^2 = 4ax, the parameters must satisfy t1t2=1t_1 t_2 = -1. If this is missed, the coordinates of QQ become incorrect. Always apply the focal chord parameter relation before using the length condition.

  • Applying the section formula incorrectly for PM:MQ=3:1PM:MQ = 3:1. The point MM is closer to QQ than to PP, so a sign or weight error gives the wrong coordinates. Use the ratio carefully and verify that the computed point lies between PP and QQ.

  • Taking the slope of the perpendicular line as 43\frac{4}{3} instead of 43-\frac{4}{3}. Perpendicular slopes are negative reciprocals, not just reciprocals. After finding slope 34\frac{3}{4} for PQPQ, the required slope must be 43-\frac{4}{3}.

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